# Proof that log 2 is an irrational number

Before doing the proof, let us recall two things: (1) rational numbers are numbers that can be expressed as $\frac{a}{b}$ where $a$ and $b$ are integers, and $b$ not equal to $0$; and (2) for any positive real number $y$, its logarithm to base $10$ is defined to be a number $x$ such that $10^x = y$. In proving the statement, we use proof by contradiction.

Theorem: log 2 is irrational

Proof:

Assuming that log 2 is a rational number. Then it can be expressed as $\frac{a}{b}$ with $a$ and $b$ are positive integers (Why?).  Then, the equation is equivalent to $2 = 10^{\frac{a}{b}}$. Raising both sides of the equation to $b$, we have $2^b = 10^a$. This implies that $2^b = 2^a5^a$.  Notice that this equation cannot hold (by the Fundamental Theorem of Arithmetic) because $2^b$ is an integer that is not divisible by 5 for any $b$, while $2^a5^a$ is divisible by 5. This means that log 2 cannot be expressed as $\frac{a}{b}$ and is therefore irrational which is what we want to show.